Figure 5 shows a simplified catapult used to hurl projectiles a long way - AQA - A-Level Physics - Question 4 - 2019 - Paper 1

To calculate the tension in the rope, we analyze the torques around the pivot.

The clockwise moment is due to the counterweight:

$ext{Clockwise Moment} = 610 ext{ kg} imes 9.81 ext{ m/s}^2 imes 1.5 ext{ m} = 8971.5 ext{ Nm}$

The anticlockwise moment due to the projectile is:

$ext{Anticlockwise Moment} = 250 ext{ N} imes 4 ext{ m} = 1000 ext{ Nm}$

Setting the moments equal gives:

T = \frac{8971.5 ext{ Nm}}{4 ext{ m}} = 2242.88 ext{ N}$$ Thus, the tension in the rope is approximately 2243 N.

Using the formula for the range of a projectile:

The range can be calculated using:

$ext{Range} = v imes t$

Where:

• $v = 18 ext{ m/s}$
• $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 7.5 ext{ m}}{9.81 ext{ m/s}^2}} \approx 1.23 ext{ s}$

So, $\text{Range} = 18 ext{ m/s} \times 1.23 ext{ s} \approx 22.14 ext{ m}$

When the sling is adjusted such that the projectile is released just before the beam is vertical, it results in a different projectile motion.

• The initial vertical velocity component is removed, potentially decreasing the time the projectile spends in the air.
• As there is no initial upward velocity, the horizontal range may be reduced compared to when it is released horizontally. In conclusion, this change generally leads to a decrease in range due to the loss of vertical velocity at the release point.